Question

If $\cot \theta =\frac{12}{5}$ show that ${\tan }^2\theta -{\sin }^2\theta ={\sin }^4\theta \times {\sec }^2\theta$

Answer 1

Let $\cot \theta =\frac{12}{5}\Rightarrow \tan \theta =\frac{5}{12},\sin \theta =\frac{5}{13},\sec \theta =\frac{13}{12}$........(1).

Now,

${\tan }^2\theta -{\sin }^2\theta$

$={\left(\frac{5}{12}\right)}^2-{\left(\frac{5}{13}\right)}^2$ [Using (1)]

$=\frac{5^4}{{12}^2{.13}^2}$.......(2).

Again

${\sin }^4\theta \times {\sec }^2\theta$

$={\left(\frac{5}{13}\right)}^4\times {\left(\frac{13}{12}\right)}^2$ [Using (1)]

$=\frac{5^4}{{12}^2{.13}^2}$.......(3)

From (2) and (3) it is clear that ${\tan }^2\theta -{\sin }^2\theta =$${\sin }^4\theta \times {\sec }^2\theta$.